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Question

(A) The electric field due to a thin infinite sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_0}$.Sin

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$2 	imes 10^{-10} \ N/C$

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(A) The electric field due to a thin infinite sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_0}$.Since both plates have the same positive surface charge density $\sigma = 17.7 	imes 10^{-22} \ C/m^2$,the electric fields produced by both plates in the outer region of the second plate point in the same direction (away from the plates).Therefore,the total electric field $E$ in the outer region is the sum of the fields from both plates:$E = E_1 + E_2 = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$.Given $\varepsilon_0 = 8.85 	imes 10^{-12} \ C^2/(N \cdot m^2)$:$E = \frac{17.7 	imes 10^{-22}}{8.85 	imes 10^{-12}} = 2 	imes 10^{-10} \ N/C$.

Two large,thin metal plates are parallel and close to each other. Their inner faces have surface charge densities of the same sign and magnitude $17.7 \times 10^{-22} \ C/m^2$. What is the electric field $E$ in the outer region of the second plate?

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